Every non-empty subset of $R$ having a lower bound has an infimum.
Since $S$ has a lower bound, then the set $S' = \{−x \mid x ∈ S\}$ has an upper bound and hence by completeness property has a supremum. Now show that if $M = \sup(S)$ then $−M = \inf(S)$.
Is there other way to prove it??
Best Answer
You seem to be assuming every set that is non-empty and bounded above has a supremum. If we assume that, then there is another way: Let $T$ be the set of all lower bounds of $-S$. Then $T$ has a supremum (since every member of $-S$ is an upper bound of $T$). Try to show that that is the infimum of $-S$ and that it is equal to $-\sup S$.